A fair n-sided die is rolled n times. Assuming the rolls are independent, calculate the probability of getting a match on roll i, i.e. on roll i the die shows i.
I don’t think you have posted what you are really asking.
Suppose we roll an ordinary die six times.
The probability of getting a 1 on the first roll is $\displaystyle \frac{1}{6}$.
The probability of getting a 2 on the second roll is $\displaystyle \frac{1}{6}$.
…
The probability of getting a 6 on the sixth roll is $\displaystyle \frac{1}{6}$.
Now is that truly what you asked?
Yes i understand that the likelihood of a matct is 1/n on any particular trial.
But i dont understand how to calulate the probability of getting a match on roll i.
Where does n start and where does it end?
If its infinite then isnt the probabilty of getting a match 1?
Suppose we roll an ordinary die six times.
If you want the probability if getting the string 123456 then that is $\displaystyle \left(\frac{1}{6}\right)^6$.
For an n-sided die rolled n times, the probability if getting the string 123…n is $\displaystyle \left(\frac{1}{n}\right)^n$.
Perhaps you mean what is the probability of getting at least one match?
If so, this is 1 - P(no match). The probability of not getting a match on the nth roll is $\displaystyle 1-1/n$, so assuming the rolls are independent, the probability of no macth in n rolls is $\displaystyle (1-1/n)^n$, and
$\displaystyle P(\text{at least one match}) = 1 - (1 - 1/n)^n$.
If you are interested in large n,
$\displaystyle \lim_{n \to \infty} 1 - (1 - 1/n)^n = 1 - 1/e$.